A natural number has exactly 10 divisers including one and itself how many distinct prime factors can these natural number have ?

• take any 2 digit number.........for ex. 20 ...we can see 20 have 6 divisors 1,2,4,5,10,20 . According to question..... number have exactly 10 divisors. So we have to choose another number which have 10 divisors .........again we have choose number 40......... we can see 40 have 8 divisors 1,2,4,5,8,10,20,40.......again we have to choose another number that is 80......we can see 80 have 10 divisors 1,2,4,5,8,10,16,20,40,80. So prime factor of number 80 is 2*2*2*2*5 and distinct prime factor of 80 is two that is 2 and 5.
  answer: There are two distinct prime factor that is 2 and 5.
  Tips : Number will be choose by doing double and calculate the divisors.
• 7 years agoHelpfull: Yes(20) No(0)
• Let N be the number and has 10 divisors.
  We know that N= a^p * b^q * c^r........
  So here a,b,c,....... are the prime factors and
  (p+1)(q+1)(r+1).....are the no. of factors.
  So number N has 10 divisors or factors which means
  N=9+1 or N=(4+1)(1+1) i.e. these 2 are the only cases.
  So which means answer is
  Either 1 or 2 solutions.
• 7 years agoHelpfull: Yes(13) No(2)
• If n is a composite number of form (a^p)*(b^q)*(c^r) where a , b , c are the prime factors of number n then number of prime factors will be given as (p+1)(q+1)(r+1)... Since here (p+1)(q+1)(r+1) = 10. Solving we get number n= (a^4)*(b) hence there can be only 2 distinct prime factors only.
• 7 years agoHelpfull: Yes(1) No(2)
• SATISH KUMAR SINGH ...sir please elaborate your answer.i cant understand.
• 7 years agoHelpfull: Yes(1) No(2)
• this que is belong to like find the number of factors of a given no. for ex:-find the no of all factor of 40?
  soln;-
  40=2^3*5^1 //then no of all factor = (3+1)(1+1)=8
  i.e 40 is divisible by total 8 different whole no. are 1,2,4,5,8,10,20,40.
  So, the conclusion is no of divisor including 1 and itself= no of all factor of any whole no.
  Now, for que given 10 divisor i.e total 10 factor.
  and, the no. of factor =10 i.e 5*2 which is also write as (4+1)(1+1)
  Lets take a no. 80 its facorization =2^4*5^1 //now see the power of the factor it is 4and 1.
  then the total no of factor = (4+1)(1+1)=10=10 divisor .
  now here the trick is no of braces() = no of distinct factor .
  Hence for 10 divisor i.e 5*2 i.e (4+1)(1+1) i.e total 2 distinct factor(then see factor is prime or not)
• 7 years agoHelpfull: Yes(1) No(0)
• As given it has 10 divisors so it can be through
  5*2 or 10*1
  so there can be maxm. of 3 prime factors and min. 2 prime factors.
• 7 years agoHelpfull: Yes(0) No(2)
• Suppose a number n is 18, it can be expressed as (2^1)*(9^2) then total no. of divisors it can have including 1 and itself is (1+1)(2+1) = 6. Similarly the above question be solved as formerly solved
• 7 years agoHelpfull: Yes(0) No(2)
• Sorry there is correction 18 = 2*(3^2)
• 7 years agoHelpfull: Yes(0) No(2)
• why not only 1 distinct prime number.suppose 512(1,2,4,8,16,32,64,128,256,512).
  plese comment..
• 7 years agoHelpfull: Yes(0) No(0)
• The number of factors or divisors will be the product of all the powers after incremented by 1 like (a+1)(a+1).(b+1)(b+1).(c+1)(c+1)..and so on.
  
  So, in the question we got 10, now 10 can be a result if either r a single power or product of 2 powers only and not more than that. So, here
  
  Case 1: (a+1)(a+1)=10 Case 2: (a+1)(a+1)(b+1)(b+1) = 10.
  
  so the answer is either 1 or 2.
• 7 years agoHelpfull: Yes(0) No(0)